The Bohr model of hydrogen

Result

The Bohr model of hydrogen

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Statement

For a one-electron atom or ion with nuclear charge \(+Ze\) (hydrogen has \(Z=1\)), postulate that the electron moves in a circular orbit under the Coulomb attraction of the nucleus, and that its orbital angular momentum is quantised in integer multiples of \(\hbar\): \(L=n\hbar\), \(n=1,2,3,\dots\). Then only discrete orbital radii and energies are allowed, with the electron's total energy given by \(E_n=-\dfrac{Z^2}{n^2}\cdot\dfrac{m_ee^4}{8\varepsilon_0^2h^2}=-\dfrac{13.6\,Z^2}{n^2}\text{ eV}\), and orbital radius \(r_n=\dfrac{n^2}{Z}a_0\), where \(a_0=\dfrac{\varepsilon_0h^2}{\pi m_ee^2}\approx0.529\,\text{Å}\) is the Bohr radius.

Why it matters

The Bohr model is the first physical picture that explains why hydrogen emits light only at specific, discrete wavelengths rather than a continuous spectrum — the empirical Rydberg formula, known since 1888 but unexplained, falls directly out of this result as the energy difference between two allowed levels. It is the bridge between the classical planetary picture of the atom (Rutherford, 1911) and full quantum mechanics: it keeps a classical orbit but imposes one quantum rule on top of it, and that single rule is enough to derive the entire hydrogen spectrum from first principles for the first time in the history of chemistry and physics.

It also fixes the scale of atomic chemistry. The Bohr radius \(a_0\) sets the size of the hydrogen atom (and, roughly, all atoms) at the Ångström scale, and the Rydberg energy \(13.6\text{ eV}\) sets the scale of chemical bond energies and ionisation energies — both numbers that recur throughout the rest of the periodic-trends story built on this unit.

Hypotheses

One electron only (hydrogen or a hydrogen-like ion such as He\(^+\), Li\(^{2+}\)).For a two-electron atom such as helium, the electrons repel each other via a Coulomb term that depends on both electrons' positions simultaneously; there is no way to solve for a single circular orbit independent of the other electron. The Bohr model gives no closed-form energy for any multi-electron atom — not even a poor approximation without further, separate assumptions (e.g. effective nuclear charge, treated later in Unit CU-101 itself). Angular momentum quantised as \(L=n\hbar\), with a circular (not elliptical) orbit.Drop the specific rule \(L=n\hbar\) and the "orbit" can take any radius classically, radiating continuously as the accelerating electron loses energy and spirals into the nucleus in about \(10^{-11}\,\)s — the classical instability problem the whole postulate exists to avoid. The correct quantum-mechanical ground state (from the Schrödinger equation) in fact has zero orbital angular momentum, \(l=0\), directly contradicting Bohr's own \(L=n\hbar\ne0\) for \(n=1\); the model gets the right energies for the wrong structural reason, addressed further in Discussion below. Non-relativistic electron.For hydrogen (\(Z=1\)) the electron's characteristic speed \(v_1=\alpha c\approx c/137\) is small enough that relativistic corrections are a fine-structure-level effect (parts in \(10^4\)–\(10^5\)). For a hydrogen-like ion with large \(Z\) (e.g. highly ionised uranium, \(Z=92\)), \(v_1=Z\alpha c\) approaches a sizeable fraction of \(c\), and the non-relativistic energy formula above becomes measurably wrong; a full treatment needs the relativistic Dirac equation.

Proof

\frac{1}{4\pi\varepsilon_0}\frac{Ze^2}{r^2} = \frac{m_ev^2}{r}

Newton's second law for uniform circular motion: the Coulomb attraction between the electron (charge \(-e\)) and the nucleus (charge \(+Ze\)) supplies exactly the centripetal force needed to keep the electron in a circular orbit of radius \(r\) and speed \(v\). A

L = m_evr = n\hbar, \qquad n=1,2,3,\dots

Bohr's quantisation postulate: the orbital angular momentum is restricted to integer multiples of \(\hbar=h/2\pi\). This is an added axiom, not derivable from classical mechanics — it is the one genuinely quantum ingredient in the whole derivation. A

v = \frac{n\hbar}{m_er} \quad\text{(from Step 2)} \ \Longrightarrow\ \frac{Ze^2}{4\pi\varepsilon_0 r^2} = \frac{m_e}{r}\left(\frac{n\hbar}{m_er}\right)^2 = \frac{n^2\hbar^2}{m_er^3}

Solve Step 2 for \(v\) and substitute into Step 1, eliminating \(v\) in favour of \(r\) and the quantum number \(n\). A

r_n = \frac{4\pi\varepsilon_0\hbar^2}{m_eZe^2}\,n^2 = \frac{n^2}{Z}\cdot\frac{4\pi\varepsilon_0\hbar^2}{m_ee^2} = \frac{n^2}{Z}a_0

Solve Step 3 for \(r\), and define the Bohr radius \(a_0\equiv4\pi\varepsilon_0\hbar^2/(m_ee^2)=\varepsilon_0h^2/(\pi m_ee^2)\) as the constant multiplying \(n^2/Z\); it is the orbital radius of the \(n=1\) state of ordinary hydrogen (\(Z=1\)). A

E = \tfrac12 m_ev^2 - \frac{Ze^2}{4\pi\varepsilon_0 r} \ \overset{\text{Step 1}}{=}\ \frac{Ze^2}{8\pi\varepsilon_0 r} - \frac{Ze^2}{4\pi\varepsilon_0 r} = -\frac{Ze^2}{8\pi\varepsilon_0 r}

Total energy is kinetic plus Coulomb potential energy. Rewrite the kinetic term using Step 1 (\(m_ev^2=Ze^2/4\pi\varepsilon_0r\)) so both terms share the same \(1/r\) dependence and combine into one term — the classical virial-theorem relation \(E=-\text{KE}\) for an inverse-square orbit falls out as a byproduct. B

E_n = -\frac{Ze^2}{8\pi\varepsilon_0}\cdot\frac{Z}{n^2a_0} = -\frac{Z^2e^2}{8\pi\varepsilon_0a_0}\cdot\frac{1}{n^2} = -\frac{Z^2}{n^2}\cdot\frac{m_ee^4}{8\varepsilon_0^2h^2}

Substitute \(r_n\) from Step 4 into Step 5, then expand \(a_0\) back out to express the prefactor purely in terms of fundamental constants. Numerically, \(m_ee^4/(8\varepsilon_0^2h^2)=13.6\,\text{eV}\), the Rydberg energy. B

Result

E_n = -\frac{13.6\,Z^2}{n^2}\text{ eV}, \qquad r_n=\frac{n^2}{Z}a_0,\qquad a_0\approx0.529\,\text{Å}

Reading. One quantisation rule (\(L=n\hbar\)) applied to a classical circular Coulomb orbit forces both the radius and the energy to jump in discrete steps indexed by the integer \(n\); larger \(n\) means a larger, less tightly bound orbit, with energy climbing toward \(0\) (ionisation) as \(n\to\infty\).

Scope. Exact for any single-electron atom or ion: H, He\(^+\), Li\(^{2+}\), and so on, with \(Z\) the nuclear charge. Not valid, even approximately without modification, for any atom with two or more electrons.

Corollaries & converses

The energy difference between two levels, \(\Delta E = 13.6\,Z^2\left(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2}\right)\text{ eV}\), combined with \(\Delta E=hc/\lambda\), is exactly the empirical Rydberg formula for the hydrogen line spectrum — the theorem this unit's next result (the Rydberg formula) makes explicit and general.
The ionisation energy of hydrogen from its ground state is \(-E_1=13.6\,\text{eV}\), the reference value against which every other atom's first ionisation energy in this unit's periodic-trends result is measured.
Converse fails: matching the correct energy levels does not mean the underlying orbital picture is correct. The full quantum-mechanical treatment (Schrödinger equation) reproduces exactly the same \(E_n=-13.6Z^2/n^2\,\text{eV}\) formula while rejecting the circular-orbit picture entirely — see Discussion.

Fails without

Drop the one-electron restriction: for helium (\(Z=2\), two electrons), the same derivation cannot be run, because Step 1 assumed a single electron feels only the nuclear Coulomb force; with a second electron present, the potential energy term must include an electron–electron repulsion \(e^2/4\pi\varepsilon_0r_{12}\) that depends on both electrons' instantaneous separation, and the two-body circular-orbit ansatz no longer solves the resulting coupled problem in closed form.
Drop the quantisation postulate (classical electrodynamics only): an accelerating charge radiates energy (Larmor's formula); a classical electron in a circular orbit continuously loses energy and its radius shrinks smoothly, predicting the atom collapses in about \(1.6\times10^{-11}\,\)s — an atom that cannot exist for more than a fraction of a nanosecond, in flat contradiction with observed matter. Quantisation of \(L\) is precisely what forbids all but a discrete set of radii, blocking the classical collapse.

Common errors

Forgetting the factor of \(Z^2\) in the energy formula (and \(1/Z\) in the radius formula) when applying the result to a hydrogen-like ion rather than to hydrogen itself — He\(^+\) has \(E_1=-13.6\times4=-54.4\,\text{eV}\), not \(-13.6\,\text{eV}\).
Treating the Bohr model as giving the correct orbital angular momentum of the electron. It does not: the true ground state has \(L=0\), while the Bohr model insists \(L=n\hbar\ge\hbar\) for every state, including \(n=1\).
Applying the model, even qualitatively, to multi-electron atoms to "explain" periodic trends directly — the model has no term for electron–electron repulsion or shielding, both essential to real periodic trends (see this unit's periodic-trends result, which uses effective nuclear charge instead).

Discussion

Niels Bohr proposed this model in 1913, three years after Rutherford's gold-foil experiment established the nuclear atom and eleven years before de Broglie's matter-wave hypothesis and thirteen years before Schrödinger's equation. Bohr's genius move was to accept the classical planetary picture at face value but bolt on a single quantisation rule borrowed from Planck's 1900 treatment of blackbody radiation, and show that this one rule reproduces the Rydberg formula — an empirical fit that had stood, unexplained, for 25 years.

The model was superseded within about a decade by full quantum mechanics, and its central image — the electron as a tiny planet on a fixed circular track — is not how electrons actually behave; there are no orbits, only probability-density orbitals (the subject of this unit's quantum-numbers-and-orbitals result). Yet the energy-level formula \(E_n=-13.6Z^2/n^2\,\text{eV}\) survives completely unchanged when the correct calculation is redone with the Schrödinger equation for the hydrogen atom — a striking case of a model reaching the right quantitative answer through an incorrect physical mechanism, because for hydrogen specifically, the accidental degeneracy of the Coulomb potential collapses the true energy dependence (on both \(n\) and the orbital quantum number \(l\)) down to a dependence on \(n\) alone.

The deeper reason the Bohr formula survives is the Coulomb potential's special "accidental" \(SO(4)\) symmetry (beyond the ordinary \(SO(3)\) rotational symmetry every central potential has), first explained via the Laplace–Runge–Lenz vector; it is this extra symmetry that forces every orbital with the same principal quantum number \(n\) (regardless of \(l\)) to share exactly the same energy, which is exactly the single-index dependence \(E_n\) that Bohr's cruder model happened to stumble onto.

Common misconception: that the Bohr model is simply "an approximation" to the true quantum picture, in the sense of being close-but-imprecise. It is not an approximation in that sense at all — its energies are exact for hydrogen — but its physical picture (electrons on fixed circular tracks) is qualitatively wrong, not merely imprecise; it predicts a definite orbital radius and momentum simultaneously, which the uncertainty principle (a later, unit-CU-101 companion topic) forbids outright.

Worked examples

\text{Hydrogen ground state: } Z=1,\ n=1 \ \Longrightarrow\ E_1=-13.6\text{ eV}, \quad r_1=a_0\approx0.529\,\text{Å}

Direct substitution into the Result. This is the reference energy scale for hydrogen: \(13.6\,\text{eV}\) is exactly the energy needed to ionise a ground-state hydrogen atom. A

\text{He}^+\text{ ion: } Z=2,\ n=1 \ \Longrightarrow\ E_1=-13.6\times2^2=-54.4\text{ eV}, \quad r_1=\tfrac12a_0\approx0.265\,\text{Å}

He\(^+\) (one electron, nuclear charge \(+2e\)) is exactly the kind of hydrogen-like ion the model covers. The orbit shrinks (extra nuclear pull) and binds four times more tightly — the \(Z^2\) scaling of the energy versus the \(1/Z\) scaling of the radius. A

\text{Lyman-}\alpha\text{ line: } n=2\to n=1,\ Z=1 \ \Longrightarrow\ \Delta E = 13.6\left(1-\tfrac14\right)=10.2\text{ eV} \ \Longrightarrow\ \lambda=\frac{hc}{\Delta E}\approx121.5\,\text{nm}

Using the corollary \(\Delta E=13.6Z^2(1/n_1^2-1/n_2^2)\,\text{eV}\) with \(n_1=1,n_2=2\), then converting to wavelength via \(\Delta E=hc/\lambda\). \(121.5\,\text{nm}\) is close to the observed wavelength of the strongest line in the hydrogen ultraviolet (Lyman) series, matching astronomical and laboratory spectroscopy to high precision. B

\lambda_{\text{Lyman-}\alpha}\approx121.5\,\text{nm}\ \ (\text{electron mass; precisely measured value: }121.567\,\text{nm})

Reading. A single input rule (\(L=n\hbar\)) reproduces a measured spectral line to high precision, which is exactly why the model was accepted immediately in 1913 despite its later-discovered conceptual problems.

Scope. The same calculation, with the appropriate \(n_1,n_2\), gives every line of the hydrogen Balmer, Lyman, Paschen and higher series — the full content of the Rydberg formula, this unit's next result.

Problems

Compute the ionisation energy of Li\(^{2+}\) (\(Z=3\)) from its ground state, and its ground-state orbital radius.
Solution
\(E_1=-13.6\times3^2=-122.4\,\text{eV}\), so the ionisation energy (energy required to remove the electron to \(r=\infty\), \(E=0\)) is \(+122.4\,\text{eV}\). The radius is \(r_1=a_0/3\approx0.176\,\text{Å}\) — smaller and more tightly bound than hydrogen or He\(^+\), continuing the \(Z^2\)/\(1/Z\) pattern of Worked Example 2.
Find the radius and speed of the electron in the \(n=4\) orbit of hydrogen, and check that the orbit's classical period is consistent with a bound (not escaping) electron.
Solution
\(r_4=4^2a_0=16a_0\approx8.46\,\text{Å}\). From Step 2, \(v=n\hbar/(m_er)\); substituting \(r_4\) and \(n=4\) gives \(v_4=\hbar/(4m_ea_0)=v_1/4\), where \(v_1=\alpha c\approx2.19\times10^6\,\text{m/s}\) is the \(n=1\) speed (a standard reference value), so \(v_4\approx5.5\times10^5\,\text{m/s}\). The orbital period \(T=2\pi r_4/v_4\) is finite and the energy \(E_4=-13.6/16=-0.85\,\text{eV}\) is negative, confirming a bound (non-escaping) orbit as expected for any finite \(n\).
A student calculates the Bohr-model energy levels for a neutral helium atom by simply setting \(Z=2\) in \(E_n=-13.6Z^2/n^2\,\text{eV}\) and comparing to the measured helium ionisation energy (\(24.6\,\text{eV}\)). Explain why this comparison is invalid, using the Hypotheses above.
Solution
Neutral helium has two electrons, not one, so the one-electron hypothesis is violated outright; the formula \(E_n=-13.6Z^2/n^2\) was derived (Step 1) assuming a single electron feels only the nuclear Coulomb attraction, with no electron–electron repulsion term. Plugging \(Z=2\) gives \(E_1=-54.4\,\text{eV}\) per electron, which is the correct energy for the hydrogen-like ion He\(^+\) (Worked Example 2) — not for neutral He, where electron–electron repulsion raises the actual first-ionisation energy to \(24.6\,\text{eV}\), roughly half the naive prediction. This is exactly the shielding effect explored properly in this unit's periodic-trends result.
Show that as \(n\to\infty\), \(E_n\to0^-\) and interpret this physically.
Solution
From the Result, \(E_n=-13.6Z^2/n^2\,\text{eV}\); as \(n\to\infty\), \(1/n^2\to0\), so \(E_n\to0\) from below (always negative for finite \(n\), approaching but never reaching zero). Physically, \(E=0\) is the threshold at which the electron is no longer bound to the nucleus (zero binding energy, infinite orbital radius, from \(r_n=n^2a_0/Z\to\infty\)); the levels \(E_n\) accumulate ever more densely just below this ionisation threshold, which is exactly what is observed spectroscopically as the Lyman/Balmer/etc. series limits, where the discrete lines converge to a continuum edge.